A man pushes a wooden crate on wooden floor. The direction of the force is 30 de

Question

A man pushes a wooden crate on wooden floor. The direction of the force is 30 degree from the horizontal, How much force does he use in order to maintain a constant speed? HOW much work is done if he pushes the crate over a distance of 25.0 m? Kinetic friction coefficient for wood/wood is 0.50. The world's largest hydropower plant, the Three Gorges Power Plant in China, has a capacity of 18,000 megawatts. How much water (in metric tons, 1 ton= 1000kg) must flow from the 120 m high dam each second to turn the water turbines so the generators can produce electricity at capacity? A 50.0 kg object slides down a 30 degree angle slope from rest at a height of 12.0 m. The coefficient of kinetic friction between the object and the slope surface is 0.20. What is the object's speed at the bottom of the slope? If the object continues to slide on the level ground after reaching the bottom, and if the ground's coefficient of friction is the same as the slope, how far will the abject move before stopping?

Explanation / Answer

12)

let mass of crate be m
hence normal reaction = mg-Fsin30 =9.8m-F/2
hence friction force =?*normal reaction =4.9m-F/4

hence for constant speed
friction force should be equal to Fcos30

hence 4.9m-0.25F =F*0.866

hence F =4.39m

work done
= Fcos30 *25 =95.04m J.............m is the mass of crate

9)

18000 megawatts = 18000 X 10^6 watts = 18 X 10^9 J/s
g= 10m/s^2
h= 120 m
PE of the water is converted to electrical energy.
so, mgh per sec = 18 X 10^9 J/s
m = 18X 10^9 / gh = 15000000 kg
= 15000 metric tonnes of water.

3)

using energy conservation,
mgh - umg*cos30*15/sin30 = .5*mv^2
so, v =sqrt(2*(9.8*15 - .25*9.8*cos30*15/sin30)) = 12.92 m/s

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